Computing Picard Lattices of K3 Surfaces

The other direction

• Two lectures of upper bounds. Reduction gave ρ≤ρ(X_p^al), and the obstruction sharpened this upper bound on the rank.
• The rank alone does not determine the Picard lattice: it says nothing about the intersection form, nothing about the Galois action, and exhibits no curve.

Today: produce the classes. A lower bound is a curve you can write down, and enough curves with their intersections is the lattice itself.

An analytic approach

H₂(X, ℤ) ⟶ℂ, γ↦∫_γω_X.

Hence, if Π∈ℂ²² represents this map, then we are reduced to finding a (saturated) lattice Λ⊂H₂(X, ℤ) of solutions

ΠR = 0, R ∈H₂(X, ℤ) ≃ℤ²².

An analytic approach

• Π can be computed:
  • rigorously as a ball via deformation for projective hypersurfaces (Sertöz), with PeriodSuite: github.com/emresertoz/PeriodSuite
  • lefschetz-family (heuristic homology basis, Lairez-Pichon-Pharabod-Vanhove): github.com/ericpipha/lefschetz-family
  • Elsenhans-Jahnel implemented also an heuristic approach for degree 2
• Heuristically, via lattice reduction algorithms, we can find Λ⊂H₂(X, ℤ).
• There is no obvious way to prove that our guesses are actually correct.

• Nonetheless, given Π as a ball, one can compute B ≫0 such that

  Pic (X)_∣B := ℤ ⟨γ∈Pic X ∣−γ²_prim < B ⟩⊆Λ (Lairez-Sert\"{o}z).

Our setup

Let C be a nice (smooth, projective, geometrically integral) curve over k of genus g given by equations. Let J be the Jacobian of C.

Heuristic solution

J = ℂ^g / Λ_J, the period lattice Λ_J computed numerically, to high precision, from a basis of H⁰(C, Ω_C).

By picking a k-basis for H⁰(C, Ω_C), we have

End (J) = { T ∈M_g(k) ∣T Λ_J ⊂Λ_J }

Hence, if Π is a period matrix for C, i.e., Λ_J = Πℤ^2g, then we are reduced to finding a ℤ-basis of the solutions (T, R) to

T Π= ΠR, T ∈M_g(k), R ∈M_2g(ℤ).

The Galois module structure of End (J) is given via its action on T ∈M_g(k).

Heuristically, via lattice reduction algorithms, we can find such a ℤ-basis.

There is no obvious way to prove that our guesses are actually correct.

Representing endomorphisms via correspondences

α_C :C ^AJ⟶ J ^α⟶ J ⇢Sym ^g(C)P ↦{Q₁, ..., Q_g} ⟺α([P − P₀]) = [ ∑_i=1^g Q_i − P₀ ]

This traces out a divisor on C ×C, which determines α.

This divisor is a certificate of containment for α∈End J.

By interpolation via α_C or by locally solving a differential equation on C ×C.

Examples

• We have verified, decomposed and matched the 6,216,959 curves over ℚ of genus 2 in the L-functions and modular form database LMFDB.org
• The algorithm verifies that the following genus 4 curve over ℚ(√(3))

  0= −8x² + 8xy + 17y² − 34xz − 2yz − 28z² − 10xw − 9yw − 18zw + 2w²,0= 4x³ − 6x² y − 6x y² + 12x² z + 6xyz + 24y² z − 12x z² − 24z³ + 2x² w + 7xyw+ 4y² w + 4xzw − 13yzw − 8z² w − 20x w² − 3z w² − 12w³

  has real multiplication by the maximal order of ℚ(x)/(x⁴ − x³ − 3x² + x + 1).

  The first step to show that, under Langlands, it corresponds to a specific Hilbert modular form f, i.e., J_ℚ(√(3)) ∼A_f. We used this in a recent project, where we show that the 2-isogeny field of A_f solves the inverse Galois problem for PSL ₂(𝔽₁₆) ⋊C₂ ≃17T7.

• Our method works just as well for isogenies and projections.
• Try it: github.com/edgarcosta/endomorphisms, putatively in Magma by the end of the semester.

Reconstructing isolated curves from their Hodge classes

Reconstructing isolated curves from their Hodge classes

In favorable circumstances we expect low order equations in I_γ to span I(C).
For example, smooth rational curves of degree up to 4 in K3s.

We need "isolated" classes, so elliptic curves are also hard. An elliptic curve C has [C]² = 0 and moves in a pencil, so its periods do not determine a single curve.

No hope to recover rational curves of degree higher than 4. For d ≥5 one needs a ≥3 before I(C)_a is non-zero, and there the Jacobian ideal contributes superfluous equations, so I(C)_a ⊊I_γ,a.

Reconstructing quadric surfaces

X : x⁴ + xyzw + y³z + yw³ + z³w = 0 ⊂ℙ³

Pic (X)|_B ⊆Λ^?⊆ Pic X

• Fortunately, there is a small Aut (Λ) orbit of size 336:
  133056 = 336 + 1008 + 1176 + 3528 ·3 + 4704 ·3 + 7056 ·9 + 14112 ·3
• For each quartic curve C ⊂X, we can compute

  I_[C],2 = ⟨a₀ x² + ⋯+ a₉ w² ⟩_ℂ

  that defines a quadric surface Q, such that Q ∩X = C ∪C. Hence, we expect an orbit of 168 quadrics each containing a pair of quartics.
• We aim reconstruct the ten (algebraic!) coefficients of these quadrics.

Reconstructing quadric surfaces

• The minimal polynomials have large height about 9k characters, e.g.:

  x¹⁶⁸ − 10014013832542203812872613924739x¹⁶¹ + 171047690745503707515328576627906817785436888130925209472262244x¹⁵⁴ − 1268317331496745879603035032448157273146519836562713924560050631153969519297207668270922371313x¹⁴⁷ + 23237703563539410755436556575134206593366430461423708193774287327245213403024087108979694756912313 ⋯

• Every computation must be done extremely selectively!
• We are presented with same 168 degree field L in 9 different ways.

Isomorphism problem

The abstract isomorphism problem feels hopeless.

In our case, we have all the compatible embeddings

σ_i : ℚ(a_k) ↪L ↪ℂ

Thus the isomorphism is given by the solution of the following linear system

{σ_i(a_k)^j}_{i, j} ·v = {σ_i(a₀)}_i, v ∈ℚ¹⁶⁸

Distinct nodes make {σ_i(a_k)^j} invertible, so the solution v ∈ℚ¹⁶⁸ is unique; the denominators of v are bounded a priori, so enough precision pins v down exactly and the isomorphism is then verified exactly.

In practice it is faster to refine the complex embeddings iteratively: their height is 4k digits, not 120k.

Intersecting the quadric surfaces with the K3 surface

Q : a₀ x² + a₁ x y + ⋯+ a₉ w² = 0 ⊂ℙ³, [L := ℚ({a_i}_i):ℚ] = 168

• It suffices to show that the singular locus S of Q ∩X consists of 10 distinct reduced points.
• Hopeless to do this directly! Operations in L are seriously expensive!
  Linear algebra. 😰 Gröbner basis. 😱
  One needs to compute S by hand, and clear denominators before that.
• Working over 𝔽_p we find 10 distinct points.
  Hence, S is zero-dimensional and reduced, and deg S ≤10.
• We conclude deg S = 10 via Gotzmann regularity theorem,
  by checking that dim L[x,y,z,w]_•/I_• = 10 for •= 6,7, where I is saturated and V(I) = S.

Computing the Galois action

Q : a₀ x² + a₁ x y + ⋯+ a₉ w² = 0 ⊂ℙ³, [L := ℚ({a_i}_i):ℚ] = 168

Q ∩X decomposes into a pair of quartics over K a quadratic extension of L.

Via the identification with the original classes we have (1)/(2 πi) ( ∫_C ω)_ω∈F¹ ∈K²¹.

These can be reconstructed in the same fashion as we reconstructed a_i.

Unclear how to certify this step! What are the denominators of (1)/(2 πi) ∫_C ω?

For Q smooth, K = L(√(disc Q)) [Costa-Sertöz].

Computing the Galois action

Q : a₀ x² + a₁ x y + ⋯+ a₉ w² = 0 ⊂ℙ³, [L := ℚ({a_i}_i):ℚ] = 168

Q ∩X decomposes into a pair of quartics over K a quadratic extension of L.

The direct computation of Gal (K/ℚ) looks hopeless.

We guess that K = F(√([)14]u) where [F : ℚ] = 24 and Gal (F/ℚ) = C₃ ×PGL (2,7).
Note, #Gal (F/ℚ) is 14 times smaller than #Aut Pic X.

There is a new paper about computing Galois groups of this kind of polynomial [Elsenhans-Steel].

1. Can we compute Gal (K/ℚ)?
2. Gal (K/ℚ) ^?= Aut Λ?
3. H¹(Gal (k/k), Pic X) = ?

Summary

We are still developing the method and figure out its applications/limitations.